Chapter 6 - The Laplace Transform - 6.1 Definition of the Laplace Transform - Problems - Page 311: 21

Converges.

Note that $\frac{d}{dx}(arctan(x))=\frac{1}{x^2+1}$. So, $$\int_0^\infty (t^2+1)^{-1} dt = \int_0^\infty \frac{dt}{1+t^2}=\lim_{x \rightarrow \infty} \arctan(t)-\arctan(0)=\pi/2$$ Therefore, the integral converges.